High Temperature Materials. By Docent. N. Menad. Luleå University of Technology ( Sweden )

Transcription

1 of Materials Course KGP003 Ch. 6 High Temperature Materials By Docent. N. Menad Dept. of Chemical Engineering and Geosciences Div. Of process metallurgy Luleå University of Technology ( Sweden )

2 Mohs scale of Hardness Minerals Mohs scale Talc 1 Gypsum 2 Calcite 3 Fluorite 4 Apatite 5 Feldspar 6 Quartz 7 Garnet Beryl (Emerald) 7.5 Topaz 8 Corund (saphire) 9 Diamant 10

3 Introduction of Materials Many materials are subjected to forces or loads during the service Ex: Al alloy ( airplane wing) The steel (in automobile axle) The mechanical behaviour of a material reflects the relationship between its response or deformation to applied load or force. The important mechanical properties are: Strengh, hardness, ductility and stiness M. P M. P are concern Laboratory experiments ( carefully ) Producers and consumers of materials, research organizations, government agencies ( interpretation of their results ) Standardization of M.P are made by professional societies In USA ASTM ( American Society for Testing and Materials )

4 Concepts of stress and strain If load is static or changes relatively slowly with time and is applied uniformly over a cross section or surface of a member, the mechanical behaviour may be performed by a simple stress-strain test. Application of load 1. Tension 2. Compression 3. Shear F F A o A o T φ l l o l o l θ F F T F A o F Tensile load produces an elongation and positive linear strain F Compressive load produces contraction and negative linear strain Tensile load produces an Shear strain γ, γ = tan θ Torsional deformation (angle twist φ) produced by an applied torque T

5 Tension Tests One of the most common mechanical stress-strain tests is performed in tension To ascertain different mechanical properties of materials ( design ) A specimen is deformed, usually to fracture, with a gradually increasing the force applied uniaxially along the long axis. See figure 6.2 The cross section is circular or rectangular The standard diameter Length of reduced section 12.8 mm (0.5 in) 4 times this diameter (60 mm) 2¼ in

6 Standard tensile specimen with circular cross section Reduced section 2¼ diameter diameter 3/4 2 Gauge length 3/8 Radius

7 This is the equipment that can be used for the tensile stress-strain Load cell The test takes several minutes The specimen is elongated by the moving crosshead Extensometer Specimen The load cell measures the magnitude of the applied load Moving crosshead The extensometer measures the elongation of specimen

8 Cement Compressive Strength Testing Compressive strength tests

9 Engineering stress The unit of σ is MPA (SI) σ = F A o F is the force applied perpendicular to the specimen cross section ( newtons (N) or pounds force (lb f ) A o Is cross section area before any force applied ( m 2 or in. 2 ) 1 MPA = 10 6 N/m 2 Engineering strain ε = l i - l o l o Δl = lo l o is the original length l i the instantaneous length Called just strain and is in dependent of the unit system, (sometimes in percentage)

10 The parameters, which are used to describe the stress-strain curve of a metal, are: Tensile strength, Yield strength or yield point, Percent elongation, Reduction of area. The first two are strength parameters; the last two indicate ductility. The engineering stress-strain curve The general shape of the engineering stress-strain curve is shown in this figure. In the elastic region stress is linearly proportional to strain. When the load exceeds a value corresponding to the yield strength, the specimen undergoes gross plastic deformation. It is permanently deformed if the load is released to zero. The stress to produce continued plastic deformation increases with increasing plastic strain, i.e., the metal strain-hardens.

11 Compression Tests exception The tests are conducted in a manner similar to the tensile test The force is compressive and the specimen contracts along the direction of stress σ = ε = l i - l o l o F A o Δl = lo Are used to compute compressive stress and strain By convention, compressive force is taken to be negative. Produces a negative stress l o > l i From The compressive strains are negative

12 Shear and Torsional Tests The shear stress τ is: A o τ = F A o F is the load or force A o is area of faces θ F F The shear strain γ is the tangent of the strain angle θ F T Torsion or a variation of pure shear, wherein a structure member is twisted in the manner of the following figure The torsional tests are performed on cylindrical solid shafts or tubes. φ T τ is a function of the applied torque T γ is related to the angle of twist φ

13 Geometric considerations of the stress state It is important to underline that the stress state is a function of the orientations of the planes See this figure Tensile stress is P P σ Applied parallel to its axis Plane oriented at some arbitrary angle θ We have a complex stress state P σ P σ τ θ Tensile stress or normal stress Shear stress σ acts to P-P plane τ acts parallel to P-P plane By using mechanics of materials principles, we can write the equations of these two stresses σ 1 + cos2θ = σ cos 2 θ = σ τ = σ sinθ cosθ = σ 2 sin2θ 2

14 Elastic deformation The behaviour of Stress-Strain unload The degree of strains depends on the magnitude of imposed stress For most metals, stress and strain are proportional to each other as given by this equation Stress Load Slope = modulus of elasticity Hooke s law σ = E ε Strain E is the modulus of elasticity or young s modulus 45 GPa (6.5 X10 6 psi) and 407 GPa (59X10 6 psi) for W Stress-strain diagram showing linear elastic deformation for loading and unloading cycles Table 6.1, p. 118 summarises the Modulus elasticity values of some metals Stress and strain are proportional Elastic deformation is non-permanent Elastic deformation When the force applied is released, the piece returns to its original shape

15 Nonlinear elastic behaviour σ Stress Δσ Δε Δσ Δε Tangent modulus = (at σ 2 ) = Secant modulus Between origin and σ 1 Strain Determination of the moduli At atomic level Small changes in the inter-atomic spacing and stretching of the inter-atomic bonds ε E α df dr ( inter-atomic forceseparation curve) r o

16 This figure shows the forceseparation curves for materials df dr r 0 Strongly bonded These materials have two inter-atomic bonds Force F 0 Weakly bonded Separation r Strong and week The for each is determined at r o

17 The modulus of elasticity for ceramic materials are higher than for metals. For polymers, they are lower Modulus of elasticity (GPa) Tungsten Steel Aluminum Temperature ( o C) These differences can be explain by the different types of atomic bonding in these materials Modulus of elasticity (10 6 psi) τ = G γ G is the shear modulus ( slope of the linear elastic region of the shear stress-strain curve These values for metals are given in Table 6.1

18 Anelasticity The elastic deformation is time dependent In most engineering materials, a timedependent elastic strain component is exist After application of the stress, the elastic deformation will continue, and upon load release some finite time is required for complete recovery This time behaviour called anelasticity For metals, the anelastic component is small (negligible) For polymers, the amount of anelastic is very high and is called viscoelastic behaviour

19 Problem 6.1 page 121 l o = 305 mm (12 in.) σ = 267 MPa E cu = 110 GPa (16X10 6 psi), from table 6.1 E cu = 110 X 10 3 MPa σ = ε E = Δ l E l o Δ l = σ l o E = 267 MPa X 305 mm 110 X 10 3 MPa Δ l = 0.77 (0.03 in.)

20 Elastic Properties of Materials The compressive strain ε x and ε y are determined If the applied stress is unixaial (in the z direction) and the material is isotropic. ε x = ε y Δl z 2 σ z lo x Δl x 2 z Poisson s ratio lo z y υ = ε x ε z = ε y ε z x For isotropic materials υ max υ should be 1/4 = 0.50 For metals and alloys υ = ( ) ( no net volume change) σ z See table 6.1 page118 For isotropic materials shear and elastic moduli are related each other E = 2G(1+υ)

21 Tensile Properties Yielding and Yielding strength Plastic deformation The elastic deformation in most structures will result by the application of stress. Elastic Plastic It is important to know the behaviour of this stress at which plastic deformation begins, or where the phenomena of yielding occurs σ y Stress P Stress σ y Upper yield point Lower yield point Strain Strain 0.002

22 Tensile Strength Typical engineering stress-strain behaviour to fracture. Point F. TS M TS is tensile strength at point M. TS may vary from 50MPa (700 psi) for Al to 300 MPa ( psi) for steels Stress Stráin Ex. Pr. 6.3 page 126

23 Ductility It is an important parameter for mechanical property Material with no plastic deformation upon fracture, called brittle See this figure Ductility of material may be expressed quantitatively (% elongation or % reduction in area) l f -l o % EL = X 100 Stress Brittle B Ductile B l o l f is fracture length and l o is the original gauge length A C Strain C % RA = A o -A f X 100 A o Tensile stress-strain behaviour and ductile materials loaded to fracture

24 Determination of ductility of materials is important because: 1. It indicates to a designer the degree to which a structure will deform plastically before fracture 2. It specifies the degree of allowable deformation during fabrication operations Brittle materials have a fracture strain < 5% Table 6.2 gives typical mechanical properties of several metals and alloys in annealed state

25 Resilience Material deformation Energy Material deformed elastically, upon unloading Absorption It is the modulus of resilience, strain energy per unit volume required to stress a material from an unload state up to point of yielding

26 The modulus of resilience for a specimen. From this figurer σ y ε y U r = σ d ε 0 Stress If the elastic region is linear U r = 1/2 σ y ε y ε y Strain σ = E ε ε σ y = U = r 1/2 σ y E σ y E = σ 2 y 2E

27 Toughness True stress and strain Ex. Pr. 6.4, 6.5 p. 133

28 Fatigue Crack Initiation

29 Strengthening/Hardening Mechanisms The ability of a crystalline material to plastically deform largely depends on the ability for dislocation to move within a material. Therefore, impeding the movement of dislocations will result in the strengthening of the material. There are a number of ways to impede dislocation movement, which include: 1. controlling the grain size (reducing continuity of atomic planes) 2. strain hardening (creating and tangling dislocations) 3. alloying (introducing point defects and more grains to pin dislocation) Strain Hardening Control of Grain Size

30 Strengthening/Hardening Mechanisms Effects of Elevated Temperature on Strain Hardened Materials When strain hardened materials are exposed to elevated temperatures, the strengthening that resulted from the plastic deformation can be lost. This can be a bad thing if the strengthening is needed to support a load. However, strengthening due to strain hardening is not always desirable, especially if the material is being heavily formed since ductility will be lowered. Heat treatment can be used to remove the effects of strain hardening. Three things can occur during heat treatment Recovery. Re-crystallization. Grain growth